Question

$log_{2} {64} + log_{2} {32} - log_{2} {16}$
find the value of this☝️
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Answer 1

$\large\underline{\sf{Solution-}}$

Given Logarithmic expression is

$\rm :\longmapsto\: log_{2}(64) + log_{2}(32) - log_{2}(16)$

can be rewritten as

$\rm \:  =  \: \: log_{2}( {2}^{6} ) + log_{2}( {2}^{5} ) - log_{2}( {2}^{4} )$

We know,

$\rm :\longmapsto\:\boxed{\tt{ log_{a}( {a}^{x} ) = x \: }}$

$\rm \:  =  \: \: 6 + 5 - 4$

$\rm \:  =  \: \: 11 - 4$

$\rm \:  =  \: 7$

Therefore,

$\red{\rm :\longmapsto\: \boxed{\sf{ log_{2}(64) + log_{2}(32) - log_{2}(16) = 7}}}$

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Additional Information

$\rm :\longmapsto\:\boxed{\tt{ log(xy) = log(x) + log(y)}}$

$\rm :\longmapsto\:\boxed{\tt{ log \frac{x}{y} = log(x) - log(y)}}$

$\rm :\longmapsto\:\boxed{\tt{ log {x}^{y} = y \: logx}}$

$\rm :\longmapsto\:\boxed{\tt{ log_{x}(x) = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ log_{ {x}^{y} }( {x}^{z} ) = \frac{z}{y} }}$

$\rm :\longmapsto\:\boxed{\tt{ log_{ {x}^{y} }( {w}^{z} ) = \frac{z}{y} log_{x}(w) }}$